Question: Find $\lim_{x\to -1}\dfrac{12x^3+12x^2}{x^4-x^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $6$ (Choice B) B $-6$ (Choice C) C $12$ (Choice D) D The limit doesn't exist
Solution: Substituting $x=-1$ into $\dfrac{12x^3+12x^2}{x^4-x^2}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{12x^3+12x^2}{x^4-x^2}$ can be simplified as $\dfrac{12}{x-1}$, for $x\neq 0,-1$. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $0$ and $-1$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{12x^3+12x^2}{x^4-x^2}=\dfrac{12}{x-1}$ for all $x$ -values in the interval $(-2,0)$ except for $x=-1$. Therefore, $\lim_{x\to -1}\dfrac{12x^3+12x^2}{x^4-x^2}=\lim_{x\to -1}\dfrac{12}{x-1}=-6$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -1}\dfrac{12x^3+12x^2}{x^4-x^2}=-6$.